Step-by-step Solution:

1. Define Probabilities: Let the probabilities of the three independent events be:

   • $P(E_1) = x$
   • $P(E_2) = y$
   • $P(E_3) = z$

   Then the probabilities of their complements are:

   • $P(E_1') = 1-x$
   • $P(E_2') = 1-y$
   • $P(E_3') = 1-z$

2. Translate Given Information into Equations: The problem provides definitions for $\alpha, \beta, \gamma,$ and $p$. Since the events are independent, we can write:

   • $\,\alpha = P(\text{only } E_1 \text{ occurs}) = P(E_1 \cap E_2' \cap E_3') = P(E_1)P(E_2')P(E_3') = x(1-y)(1-z)$
   • $\beta = P(\text{only } E_2 \text{ occurs}) = P(E_1' \cap E_2 \cap E_3') = P(E_1')P(E_2)P(E_3') = (1-x)y(1-z)$
   • $\gamma = P(\text{only } E_3 \text{ occurs}) = P(E_1' \cap E_2' \cap E_3) = P(E_1')P(E_2')P(E_3) = (1-x)(1-y)z$
   • $p = P(\text{none occur}) = P(E_1' \cap E_2' \cap E_3') = P(E_1')P(E_2')P(E_3') = (1-x)(1-y)(1-z)$

3. Analyze the First Given Equation: The first equation is $(\alpha - 2\beta)p = \alpha \beta$. Since all probabilities are in $(0, 1)$, we know $\alpha, \beta, p$ are non-zero. We can divide by $\alpha\beta p$: $\frac{\alpha - 2\beta}{\alpha\beta} = \frac{1}{p}$ $\frac{1}{\beta} - \frac{2}{\alpha} = \frac{1}{p}$ Now, substitute the expressions from Step 2: $\frac{1}{(1-x)y(1-z)} - \frac{2}{x(1-y)(1-z)} = \frac{1}{(1-x)(1-y)(1-z)}$ To clear the denominators, multiply the entire equation by $(1-x)(1-y)(1-z)xy$: $x(1-y) - 2y(1-x) = xy$ $x - xy - 2y + 2xy = xy$ $x - 2y + xy = xy$ $x - 2y = 0 \implies x = 2y$

4. Analyze the Second Given Equation: The second equation is $(\beta - 3\gamma)p = 2\beta\gamma$. Again, we can divide by $2\beta\gamma p$: $\frac{\beta - 3\gamma}{2\beta\gamma} = \frac{1}{p}$ $\frac{1}{2\gamma} - \frac{3}{2\beta} = \frac{1}{p}$ Substitute the expressions from Step 2: $\frac{1}{2(1-x)(1-y)z} - \frac{3}{2(1-x)y(1-z)} = \frac{1}{(1-x)(1-y)(1-z)}$ Multiply the entire equation by $2(1-x)(1-y)(1-z)yz$ to clear the denominators: $y(1-z) - 3z(1-y) = 2yz$ $y - yz - 3z + 3yz = 2yz$ $y - 3z + 2yz = 2yz$ $y - 3z = 0 \implies y = 3z$

5. Combine the Results: From Step 3, we found $x = 2y$. From Step 4, we found $y = 3z$. Substituting the second result into the first, we get: $x = 2(3z) = 6z$

6. Calculate the Required Ratio: The question asks for the value of $\frac{\text{Probability of occurrence of } E_1}{\text{Probability of occurrence of } E_3}$. This ratio is $\frac{P(E_1)}{P(E_3)} = \frac{x}{z}$. Using the relationship from Step 5, $x=6z$, we have: $\frac{x}{z} = \frac{6z}{z} = 6$ The problem states that all probabilities are in $(0,1)$. Our result is consistent with this. For example, if $z=0.1$, then $y=0.3$ and $x=0.6$, all of which are valid probabilities.

Final Answer: The required value is 6.